1. Field of the Invention
The present invention relates generally to a transmission/reception apparatus and a method for packet retransmission in a CDMA (Code Division Multiple Access) mobile communication system, and in particular, to a transmission/reception apparatus and a method for performing packet retransmission according to a channel environment.
2. Description of the Related Art
These days, mobile communication systems have evolved from an early voice-based communication system into a high-speed, high-quality radio data packet communication system for providing a data service and a multimedia service. In addition, a 3rd generation mobile communication system, divided into an asynchronous 3GPP (3rd Generation Partnership Project) system and a synchronous 3GPP2 (3rd Generation Partnership Project 2) system, is being standardized for a high-speed, high quality radio data packet service. For example, standardization on HSDPA (High-Speed Downlink Packet Access) is performed by the 3GPP, while standardization on 1xEV-DV is performed by the 3GPP2. Such standardizations are carried out to find out a solution for a high-speed, high-quality radio data packet transmission service of 2 Mbps or over in the 3rd generation mobile communication system. Meanwhile, a 4th generation mobile communication system is proposed to provide a high-speed, high-quality multimedia service superior to that of the 3rd generation mobile communication system.
In a mobile communication system providing such a communication service, a principal factor of impeding the high-speed, high-quality radio data service lies in the radio channel environment. The radio channel environment frequently changes due to a variation in signal power caused by white nose and fading, shadowing, Doppler effect caused by a movement of and a frequent change in speed of a UE (User Equipment), and interference caused by other users and a multipath signal. Therefore, in order to provide the high-speed radio data packet service, there is a need for an improved technology capable of increasing adaptability to the variation in the channel environment in addition to the general technology provided for the existing 2nd or 3rd generation mobile communication system. A high-speed power control method used in the existing system also increases adaptability to the variation in the channel environment. However, both the 3GPP and the 3GPP2, carrying out standardization on the high-speed data packet transmission, make reference to AMCS (Adaptive Modulation/Coding Scheme) and HARQ (Hybrid Automatic Repeat Request).
AMCS is a technique for adaptively changing a modulation type and a coding rate of a channel encoder according to a variation in the downlink channel environment. Commonly, to detect the downlink channel environment, a UE measures a signal-to-noise ratio (SNR) and transmits the SNR information to a Node B over an uplink. The Node B predicts the downlink channel environment based on the SNR information, and designates proper modulation type and coding rate according to the predicted value. The HSDPA and 1xEV-DV consider using the modulations of QPSK (Quadrature Phase Shift Keying), 8PSK (8-ary Phase Shift Keying), 16QAM (16-ary Quadrature Amplitude Modulation) and 64QAM (64-ary Quadrature Amplitude Modulation), and the coding rates of ½ and ¾. Therefore, an AMCS system applies the high-order modulations (16QAM and 64QAM) and the high coding rate ¾ to the UE located in the vicinity of the Node B, having a good channel environment, and applies the low-order modulations (QPSK and 8PSK) and the low coding rate ½ to the UE located in a cell boundary. In addition, compared with the existing high-speed power control method, AMCS decreases an interference signal, thereby improving the average system performance.
HARQ is a link control technique for correcting an error by retransmitting the errored data upon occurrence of a packet error at initial transmission. Generally, HARQ is classified into Chase Combining (CC), Full Incremental Redundancy (FIR), and Partial Incremental Redundancy (PIR).
CC is a technique for transmitting a packet such that the whole packet transmitted at retransmission is equal to the packet transmitted at initial transmission. In this technique, a receiver combines the retransmitted packet with the initially transmitted packet previously stored in a buffer thereof by a predetermined method. By doing so, it is possible to increase reliability of coded bits input to a decoder, thus resulting in an increase in the system performance. Combining the two same packets is similar to repeated coding in terms of effects, so it is possible to attain a performance gain of about 3 dB on the average.
FIR is a technique for retransmitting a packet comprised of only the parity bits generated from the channel encoder instead of the initially transmitted packet, thus increasing a decoding gain of a decoder in the receiver. That is, the decoder uses the new parity bits as well as the initially transmitted information resulting in a decrease in the coding rate, thereby improving performance of the decoder. It is well known in coding theory that a performance gain of a low coding rate is higher than that of repeated coding. Therefore, FIR is superior to CC in terms of the performance gain.
Unlike FIR, PIR is a technique for transmitting a combined data packet of the systematic bits and the new parity bits at retransmission. Therefore, PIR can obtain the similar effect to CC by combining the retransmitted systematic bits with the initially transmitted systematic bits during decoding, and also obtain the similar effect to the FIR by performing the decoding using the parity bits. PIR has a coding rate slightly higher than that of the FIR, showing performance between FIR and CC. However, HARQ should be considered in the light of not only the performance, but also the system complexity such as a buffer size and signaling load of the receiver, so it is not easy to determine by only one of them.
AMCS and HARQ are separate techniques for increasing adaptability to the variation in the link environment. Preferably, it is possible to remarkably improve the system performance by combining the two techniques. That is, the transmitter determines a modulation type and a coding rate that is proper for a downlink channel condition by AMCS, and then transmits packet data according to the determined modulation type and coding rate, and the receiver requests a retransmission upon failure to decode the data packet transmitted by the transmitter. Upon receipt of the retransmission request from the receiver, the Node B retransmits the data packet by HARQ technique.
FIG. 1 illustrates a transmitter in a conventional mobile communication system for high-speed packet data transmission, wherein various AMCSs and HARQs can be realized by controlling a channel encoder 112 and a modulator 118.
Referring to FIG. 1, the channel encoder 112 is comprised of an encoder and a puncturer. When input data that is proper to a data rate is applied to an input terminal of the channel encoder 112, the encoder performs encoding in order to decrease a transmission error rate. The puncturer performs puncturing on the coded bits from the encoder according to a puncturing pattern. The puncturing pattern is provided from a puncturing pattern selector 120 according to a coding rate and a modulation order previously determined by a controller 122. The coded bits punctured by the puncturer are serially provided to an interleaver 116. The interleaver 116 interleaves the punctured coded bits. The interleaver 116, a device for coping with fading that occurs in a radio channel, disperses bits constituting one information word (e.g., one word of a voice signal) thereby to decrease a probability that one information word will be lost at the same time. The interleaved signal by the interleaver 116 is modulated by the modulator 118 by a given symbol mapping method, and transmitted over a radio channel. The symbol mapping method performed in the modulator 118 is determined according to a modulation type previously determined by the controller 122. Further, a rate matcher 114 is illustrated in FIG. 1. The rate matcher 114 performs rate matching to the number of bits transmitted over a physical channel by performing puncturing or repetition on systematic bits and parity bits provided from the channel encoder 112.
A receiver performs error check on a packet received from the transmitter and informs the transmitter of the error check result. If there is no error, the transmitter transmits a new packet. However, if there is an error, the transmitter retransmits the previously transmitted packet. Retransmission of the previously transmitted data is performed by one of the above-stated HARQ types. According to the HARQ type used for the retransmission, the receiver uses the same puncturing pattern (for the CC) or selects a new puncturing pattern (for the FIR or PIR).
A detailed structure of a turbo encoder used as the channel encoder 112 of FIG. 1 is illustrated in FIG. 2. Referring to FIG. 2, the channel encoder 112 includes encoders 212 and 214 with a mother coding rate ⅙, and a puncturer 216. It is well known that a channel coding technique using the turbo encoder shows performance closets to the Shannon limit in terms of a bit error rate (BER) even at a low SNR. Therefore, in the 3GPP and 3GPP2 carrying out standardization on the future mobile communication system for high-speed multimedia data transmission with high reliability, the turbo encoder is adopted as a standard channel encoder of the HSDPA and the 1xEV-DV.
Outputs of the encoders 212 and 214 are divided into systematic bits and parity bits. The “systematic bits” refer to actual systematic bits to be transmitted, while the “parity bits” refer to a signal added to help the receiver correct a possible transmission error. The puncturer 216 selectively punctures the systematic bits or the parity bits according to a puncturing pattern selected by a puncturing pattern selector 120, thus satisfying a predetermined coding rate and modulation type.
Referring to FIG. 2, an input signal is output as a systematic bit X and, at the same time, provided to the first channel encoder 212. The first encoder 212 encodes the input signal and outputs two different parity bits Y1 and Y2. Further, an interleaver 210 interleaves the input signal. The interleaved signal is output as an interleaved systematic bit X′, and at the same time, provided to the second encoder 214. The second encoder 214 encodes the interleaved signal and outputs two different parity bits Z1 and Z2. Coded bits including the systematic bits X and X′, and the parity bits Y1, Y2, Z1, and Z2 are provided to the puncturer 216. The puncturer 216 punctures the coded bits using a puncturing pattern selected by the puncturing pattern selector 120 according to a control signal from the controller 122 or a retransmission request signal, and outputs desired systematic bits and parity bits.
The puncturing pattern used by the puncturer 216 is provided from the puncturing pattern selector 120. The puncturing pattern depends upon the coding rate and the HARQ. That is, in the case of the CC, it is possible to transmit the same packet at both initial transmission and retransmission by puncturing the coded bits such that the puncturer 216 has a fixed combination of the systematic bits and the parity bits according to a given coding rate. In the case of the FIR and PIR, the puncturer 216 punctures the coded bits in a combination of the systematic bits and the parity bits according to the given coding rate at initial transmission, and punctures the coded bits in a different combination at each retransmission. In the case of both the PIR and the FIR, the puncturing is performed in various combinations of the parity bits, resulting in an increase in the overall coding rate.
Puncturing pattern matrixes that can be selected by the puncturing pattern selector 120 according to a combination of the HARQ (CC, PIR, and FIR) and the coding rate (½ and ¾) are shown by the following formulas. In the following formulas, Pi indicates a puncturing pattern matrix used at ith retransmission including initial transmission. In the puncturing pattern, “1” indicates a non-punctured transmission bit, and “0” indicates a punctured bit. The input bits sequentially repeatedly use the puncturing pattern from the left column to the right column. A size of a row of the puncturing pattern is determined according to a mother coding rate (⅙ is used herein), and a size of the column is determined according to intention of the designer.
Of the following formulas, Equation (1) to (3) show puncturing pattern matrixes for the CC, the PIR and the FIR at the coding rate ½, respectively. In the puncturing pattern matrixes shown in Equation (1) to (3), each column includes two 1's, and one parity bit is transmitted for each systematic bit X.
Of the following formulas, Equation (4) to (6) show puncturing pattern matrixes for the CC, the PIR and the FIR at the coding rate ¾, respectively. The puncturing pattern matrixes shown in Equation (4) to (6), are defined such that one parity bit is transmitted for every three systematic bits X. However, in the case of the FIR, only parity bits are transmitted at retransmission, so 1's in the puncturing pattern matrix shown in Equation (6) are all assigned to the parity bits.
Equation (1) shows puncturing pattern matrixes for the CC HARQ at the coding rate ½.
                                                                        P                1                            =                              [                                                                            1                                                              1                                                                                                  1                                                              0                                                                                                  0                                                              0                                                                                                  0                                                              0                                                                                                  0                                                              0                                                                                                  0                                                              1                                                                      ]                                                                                                        P                2                            =                              [                                                                            1                                                              1                                                                                                  1                                                              0                                                                                                  0                                                              0                                                                                                  0                                                              0                                                                                                  0                                                              0                                                                                                  0                                                              1                                                                      ]                                                                                                        P                3                            =                              [                                                                            1                                                              1                                                                                                  1                                                              0                                                                                                  0                                                              0                                                                                                  0                                                              0                                                                                                  0                                                              0                                                                                                  0                                                              1                                                                      ]                                                                                                        P                4                            =                              [                                                                            1                                                              1                                                                                                  1                                                              0                                                                                                  0                                                              0                                                                                                  0                                                              0                                                                                                  0                                                              0                                                                                                  0                                                              1                                                                      ]                                                                        Equation  (1)            
Equation (2) shows puncturing pattern matrixes for the PIR HARQ at the coding rate ½.
                                          P            1                    =                      [                                                  ⁢                                                            1                                                  1                                                                              1                                                  0                                                                              0                                                  0                                                                              0                                                  0                                                                              0                                                  0                                                                              0                                                  1                                                      ⁢                                                  ]                          ⁢                                  ⁢                              P            2                    =                      [                                                  ⁢                                                            1                                                  1                                                                              1                                                  0                                                                              0                                                  0                                                                              0                                                  0                                                                              0                                                  1                                                                              0                                                  0                                                      ⁢                                                  ]                          ⁢                                  ⁢                              P            3                    =                      [                                                  ⁢                                                            1                                                  1                                                                              0                                                  0                                                                              1                                                  0                                                                              0                                                  0                                                                              0                                                  0                                                                              0                                                  1                                                      ⁢                                                  ]                          ⁢                                  ⁢                              P            4                    =                      [                                                  ⁢                                                            1                                                  1                                                                              0                                                  0                                                                              1                                                  0                                                                              0                                                  0                                                                              0                                                  1                                                                              0                                                  0                                                      ⁢                                                  ]                                              Equation        ⁢                                  ⁢                  (          2          )                    
Equation (3) shows puncturing pattern matrixes for the FIR HARQ at the coding rate ½.
                                          P            1                    =                      [                                                  ⁢                                                            1                                                  1                                                                              1                                                  0                                                                              0                                                  0                                                                              0                                                  0                                                                              0                                                  0                                                                              0                                                  1                                                      ⁢                                                  ]                          ⁢                                  ⁢                              P            2                    =                      [                                                  ⁢                                                            0                                                  0                                                                              1                                                  0                                                                              1                                                  0                                                                              0                                                  0                                                                              0                                                  1                                                                              0                                                  1                                                      ⁢                                                  ]                          ⁢                                  ⁢                              P            3                    =                      [                                                  ⁢                                                            0                                                  0                                                                              0                                                  1                                                                              0                                                  1                                                                              0                                                  0                                                                              1                                                  0                                                                              1                                                  0                                                      ⁢                                                  ]                          ⁢                                  ⁢                              P            4                    =                      [                                                  ⁢                                                            0                                                  0                                                                              1                                                  0                                                                              0                                                  1                                                                              0                                                  0                                                                              1                                                  0                                                                              0                                                  1                                                      ⁢                                                  ]                                              Equation  (3)            
Equation (4) shows puncturing pattern matrixes for the CC HARQ at the coding rate ¾.
                                                                        P                1                            =                              [                                                                            1                                                              1                                                              1                                                              1                                                              1                                                              1                                                                                                  0                                                              0                                                              1                                                              0                                                              0                                                              0                                                                                                  0                                                              0                                                              0                                                              0                                                              0                                                              0                                                                                                  0                                                              0                                                              0                                                              0                                                              0                                                              0                                                                                                  0                                                              0                                                              0                                                              0                                                              0                                                              1                                                                                                  0                                                              0                                                              0                                                              0                                                              0                                                              0                                                                      ]                                                                                                        P                2                            =                              [                                                                            1                                                              1                                                              1                                                              1                                                              1                                                              1                                                                                                  0                                                              0                                                              1                                                              0                                                              0                                                              0                                                                                                  0                                                              0                                                              0                                                              0                                                              0                                                              0                                                                                                  0                                                              0                                                              0                                                              0                                                              0                                                              0                                                                                                  0                                                              0                                                              0                                                              0                                                              0                                                              1                                                                                                  0                                                              0                                                              0                                                              0                                                              0                                                              0                                                                      ]                                                                                                        P                3                            =                              [                                                                            1                                                              1                                                              1                                                              1                                                              1                                                              1                                                                                                  0                                                              0                                                              1                                                              0                                                              0                                                              0                                                                                                  0                                                              0                                                              0                                                              0                                                              0                                                              0                                                                                                  0                                                              0                                                              0                                                              0                                                              0                                                              0                                                                                                  0                                                              0                                                              0                                                              0                                                              0                                                              1                                                                                                  0                                                              0                                                              0                                                              0                                                              0                                                              0                                                                      ]                                                                        Equation  (4)            
Equation (5) shows puncturing pattern matrixes for the PIR HARQ at the coding rate ¾.
                                                                        P                1                            =                              [                                                                            1                                                              1                                                              1                                                              1                                                              1                                                              1                                                                                                  0                                                              0                                                              1                                                              0                                                              0                                                              0                                                                                                  0                                                              0                                                              0                                                              0                                                              0                                                              0                                                                                                  0                                                              0                                                              0                                                              0                                                              0                                                              0                                                                                                  0                                                              0                                                              0                                                              0                                                              0                                                              1                                                                                                  0                                                              0                                                              0                                                              0                                                              0                                                              0                                                                      ]                                                                                                        P                2                            =                              [                                                                            1                                                              1                                                              1                                                              1                                                              1                                                              1                                                                                                  0                                                              0                                                              0                                                              0                                                              0                                                              0                                                                                                  0                                                              0                                                              1                                                              0                                                              0                                                              0                                                                                                  0                                                              0                                                              0                                                              0                                                              0                                                              0                                                                                                  0                                                              0                                                              0                                                              0                                                              0                                                              0                                                                                                  0                                                              0                                                              0                                                              0                                                              0                                                              1                                                                      ]                                                                                                        P                3                            =                              [                                                                            1                                                              1                                                              1                                                              1                                                              1                                                              1                                                                                                  0                                                              0                                                              0                                                              0                                                              0                                                              0                                                                                                  0                                                              0                                                              0                                                              0                                                              0                                                              1                                                                                                  0                                                              0                                                              0                                                              0                                                              0                                                              0                                                                                                  0                                                              0                                                              1                                                              0                                                              0                                                              0                                                                                                  0                                                              0                                                              0                                                              0                                                              0                                                              0                                                                      ]                                                                        Equation  (5)            
Equation (6) shows puncturing pattern matrixes for the FIR HARQ at the coding rate ¾.
                                                                        P                1                            =                              [                                                                            1                                                              1                                                              1                                                              1                                                              1                                                              1                                                                                                  0                                                              0                                                              1                                                              0                                                              0                                                              0                                                                                                  0                                                              0                                                              0                                                              0                                                              0                                                              0                                                                                                  0                                                              0                                                              0                                                              0                                                              0                                                              0                                                                                                  0                                                              0                                                              0                                                              0                                                              0                                                              1                                                                                                  0                                                              0                                                              0                                                              0                                                              0                                                              0                                                                      ]                                                                                                        P                2                            =                              [                                                                            0                                                              0                                                              0                                                              0                                                              0                                                              0                                                                                                  1                                                              0                                                              0                                                              0                                                              1                                                              0                                                                                                  0                                                              1                                                              0                                                              1                                                              0                                                              0                                                                                                  0                                                              0                                                              0                                                              0                                                              0                                                              0                                                                                                  0                                                              0                                                              1                                                              0                                                              0                                                              1                                                                                                  0                                                              0                                                              1                                                              0                                                              0                                                              1                                                                      ]                                                                                                        P                3                            =                              [                                                                            0                                                              0                                                              0                                                              0                                                              0                                                              0                                                                                                  0                                                              0                                                              1                                                              0                                                              0                                                              1                                                                                                  0                                                              0                                                              1                                                              0                                                              0                                                              1                                                                                                  0                                                              0                                                              0                                                              0                                                              0                                                              0                                                                                                  1                                                              0                                                              0                                                              0                                                              1                                                              0                                                                                                  0                                                              1                                                              0                                                              1                                                              0                                                              0                                                                      ]                                                                        Equation  (6)            
As shown in Equation (1) and Equation (4), in the case of the CC, the same puncturing pattern matrix is used regardless of the number of transmissions. As shown in Equation (2) and Equation (5), in the case of the PIR, the puncturing pattern matrix for transmitting the same systematic bits at both initial transmission and retransmission and transmitting the parity bits in various combinations is used. As shown in Equation (3) and Equation (6), the case of the FIR, the puncturing pattern matrix for not transmitting the systematic bits from ith (i≧2) retransmission and transmitting only the parity bits in various combinations is used.
As described above, the HARQ and the AMCS have contributed to an improvement in the overall system performance for high-speed packet communication. In addition, many attempts are being made to improve these techniques. For example, a technique for changing an AMCS rate in response to a variation in a condition of a transmission channel during retransmission has been proposed. However, in the AMCS rate changing technique, a modulation order and a coding rate are separately changed, so an amount of transmission data cannot coincide with TTI (Transmission Time Interval), a fixed transmission unit. Therefore, the TTI should vary according to the varying modulation order and coding rate. Since the variation in the TTI causes a considerable increase in complexity of a transceiver, it is difficult to realize this technique. For example, in order to use different TTIs at initial transmission and retransmission, supplemental information on a length of the TTIs should be transmitted from a transmitter to a receiver, or the receiver should have the information. In addition, the receiver should have a memory designed to coincide with the maximum length regardless of the length of the varying TTI.
Meanwhile, using the channel information determined for initial transmission again at retransmission as described above may become a major factor in decreasing the system performance. For example, retransmission is requested when initial transmission is failed. The failure of initial transmission occurs because the channel information estimated at initial transmission is incorrect or the channel environment has been changed during a delay time from the estimation until actual transmission. Since it is difficult to correctly estimate the channel environment and the channel environment is subject to a frequent change, maintaining the channel information estimated for initial transmission during transmission may become a factor in decreasing the overall system efficiency.